New answers tagged collision
2
This problem has a recursive flavor that we'll not try to avoid.
Conservation of momentum tells us that
$$m v_0 + (p+n-1)m v(n-1) = (p+n)m v(n).$$
Imposing the boundary condition $v(0)=0$ we find
$$v(n) = \frac{n}{n+p}v_0$$
as claimed.
Let $a_n$ be the time at which the $n$th bullet strike occurs.
We have $a_1=x_0/v_0$ and
$$v_0 (a_n - T) = v_0 ...
-2
The time taken between the N-1 collision and the N collision is $T-T\frac{N-1}{p}=T\frac{p-1+N}{p}$
Edit:
Reasoning:
The difference in T is due to the N-1 collision and is given by: $T\frac{N-1}{p}$
0
This equation works but for those components of velocities in direction of contact of two bodies i.e in the direction of forces they exert on each other,in the direction perpendicular to the force the velocities won't change.
0
Let's try just considering what could happen based on conservation laws. The two electrons have a charge of -2e, so the end product must as well. Lepton number conservation is required also, and we have $L_e=2$ here. At this level, it looks difficult to produce additional particles which satisfy just these two conservation laws. If you work in QED the only ...
4
As far as I know, nobody has ever done this, at least not at what we currently consider high energy. (Electron-electron collisions happen at low energy all the time, of course.) I doubt that anything interesting would happen, primarily because electrons are mutually repulsive, and they have a low mass. That means two colliding electrons would just bounce ...
1
You're not doing anything wrong, the objects will have different momenta in different reference frames. What should be the same in every reference frame is the forces acting on the objects during the collision. The laws of physics are the same in every reference frame, but not necessarily the numbers that go into the equations.
By way of example, lets ...
1
The problem is that your frame of reference, if you put it in an object that is accelerating is not an inertial one; the discrepancies are due to inertial forces that you're not taking into account. Why don't you just observe things from a fixed, absolute, inertial reference frame, instead?
2
Without friction, the forces during the collision (glancing or head-on) are applied exclusively through their centres of mass. (Illustration available on Wikipedia.)
The torque is given by $\tau=\mathbf r \times \mathbf F$ - but if the forces are applied through the centre of mass, then $\mathbf r$ and $\mathbf F$ are parallel, and hence $\tau=0$.
Without ...
0
With your current assumptions you do not have enough equations to solve this problem, since it is two dimensional, which gives you 4 unknown variables: $u_{1}',v_{1}',u_{1}',v_{2}'$ where $u$ and $v$ are the speeds in the respectively $x$ and $y$ direction, the indexes $1$ and $2$ indicate which object it is and the apostrophe ($'$) indicates that these ...
0
Here's a formula to help you find final velocities from initial velocities:
$$v_1=\frac{u_1(m_1-m_2)+2m_2u_2}{m_1+m_2}$$
$$v_2=\frac{u_2(m_2-m_1)+2m_1u_1}{m_1+m_2}$$
These formulas you get from combining momentum and energy equations. You have to apply both of the above formulas separately in 2 dimensions: $x$ and $y$.
So you should get ...
2
Feeling silly now. Just equating the component of velocities along the wall: $$1/2 sin\theta=cos\theta$$ we get $$\tan\theta =2$$ so, $$e=1/4$$
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